SME
Transportation Engineering · Problem type
Webster’s optimum cycle
Use this for “A signal loses L = 12 s each cycle and its critical flow ratios add to Y = 0.6. Find Webster’s optimum cycle.”
Chart
Effective green: 45.5 s
Lost time: 12 s
This is an example. Type your own values to start a new problem.
C₀ (s)
=
(1.5
×
L (s)
+
5)
÷
(1
−
Y
)
G (s)
=
C₀ (s)
−
L (s)
Formulas
C₀
=
1.5 ×
L
+ 5
1 −
Y
57.5
=
1.5 × 12 + 5
1 − 0.6
G
=
C₀
−
L
45.5 = 57.5 − 12
Assumptions
•
Webster’s cycle gives the least average delay for a pretimed signal with steady arrivals.
•
L is the lost time of a whole cycle: the start-up and clearance of every phase.
•
Y adds each phase’s critical flow ratio, flow over saturation flow; it must be below 1.
•
The effective green G = C₀ − L is shared by the phases in proportion to their ratios.
Step-by-step
WE KNOW
L = 12 s, Y = 0.6
FIND
optimum cycle (C₀), effective green a cycle (G)
Step 1 · Find optimum cycle (C₀)
Use C₀ = (1.5 × L + 5) ÷ (1 − Y)
Webster’s
rule:
more
lost
time,
or
a
busier
junction
(
Y
near
1),
needs
a
longer
cycle.
C₀
=
1.5 × 12 + 5
1 − 0.6
C₀
=
18 + 5
0.4
C₀
=
23
0.4
C₀
=
57.5
s
Step 2 · Find effective green a cycle (G)
Use G = C₀ − L
Whatever of the cycle is not lost is green the phases can use.
G
=
57.5
−
12
G
=
45.5
s
Check
Put the numbers back in. Both sides should match.
57.5
=
1.5 × 12 + 5
1 − 0.6
✓
45.5 = 57.5 − 12 ✓
Topic check
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